Determine the locus of the point $\displaystyle O$ of intersection of the altitudes (orthocenter) of a triangle $\displaystyle ABC$, if the locus of the vertex $\displaystyle A$ is a line parallel to $\displaystyle BC$.
The locus of the orthocenter is a parabola passing through B and C.
Let $\displaystyle \alpha$ denote the interior angle at A, then
1. the parabola cuts the parallel to BC through A in 2 points if $\displaystyle \alpha > 90^\circ$;
2. the parallel to BC through A is tangent to the parabola if $\displaystyle \alpha = 90^\circ$;
3. the vertex of the parabola is located between the parallel to Bc and the line BC if $\displaystyle \alpha < 90^\circ$;
Second attempt:
Place the base BC of the triangle on the x-axis. Let d denote the distance between the 2 parallels. B(0,0) and C(c,0) and A(x,d).
The slope of AC is $\displaystyle m_{AC}=-\frac d{c-x}$. Then the height through B (perpendicular to AC) has the slope:
$\displaystyle m_{h_B}=\frac{c-x}d$
Then the orthocenter is $\displaystyle H \left( x,\frac{(c-x)x}d\right)$
That means all orthocenters lie on the curve of
$\displaystyle y=-\frac1dx^2+\frac cdx$
I've choosen d = 5, C(8,0)